Consider the equation for the combustion of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), the main ingredient in nail polish remover: $$ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) } \\ {\Delta H_{\text {nxn }}=-1790 \mathrm{~kJ}} \end{array} $$ If a bottle of nail polish remover contains \(155 \mathrm{~g}\) of acetone, how much heat is released by its complete combustion?

Short Answer

Expert verified
The complete combustion of 155 g of acetone releases 4763 kJ of heat.

Step by step solution

01

Calculate the number of moles of acetone

To find the amount of heat released by the combustion of acetone, start by determining the number of moles of acetone. We can use the molar mass to convert grams to moles. The molar mass of acetone \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}\) is calculated as follows: \(3 \times 12.01 \, \text{g/mol}\) for carbon, \(6 \times 1.008 \, \text{g/mol}\) for hydrogen, and \(1 \times 16.00 \, \text{g/mol}\) for oxygen, adding up to \(3(12.01) + 6(1.008) + 16.00 = 58.08 \, \text{g/mol}\). Therefore, the number of moles of acetone is found using the formula: \(\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\).
02

Calculate the heat released per mole of acetone

The balanced chemical equation indicates that the combustion of 1 mole of acetone releases \(1790 \, \text{kJ}\) of heat. Thus, the heat released per mole can be used to calculate the total amount of heat released for our given mass of acetone.
03

Calculate the total heat released

Now we can calculate the total heat released by multiplying the number of moles of acetone by the heat released per mole, according to the equation: \(\text{total heat released} = \text{number of moles} \times \Delta H_{\text{nxn}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemistry is full of different kind of events, but few are as thrilling as chemical reactions. These are processes in which one set of chemicals (the reactants) is transformed into a new set of chemicals (the products). For students, understanding a chemical reaction begins with the balanced chemical equation. In our case, we're examining the combustion of acetone, a common solvent in nail polish remover.

Combustion is a specific type of chemical reaction that typically involves a substance reacting rapidly with oxygen and releasing energy in the form of heat or light. For the combustion of acetone \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}\), we write the balanced equation as follows:
\[\mathrm{C}_3\mathrm{H}_6\mathrm{O}(l) + 4\mathrm{O}_2(g) \longrightarrow 3\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(g)\]
This equation tells us that liquid acetone reacts with oxygen gas to produce carbon dioxide and water vapor, both in gaseous form. The balancing of the equation is crucial, as it obeys the law of conservation of mass, stating that matter is neither created nor destroyed in a chemical reaction. Hence, for each atom type, the same number of atoms must appear on both sides of the equation.
Stoichiometry
Once you've gotten to grips with the balanced chemical equation, stoichiometry is your next challenge. This is the section of the chemistry rulebook that tells you about the quantities of reactants and products involved in a chemical reaction. It's a little bit like a recipe, indicating how much of each ingredient you need and what quantity of product you'll end up with.

In the reaction of acetone, we use stoichiometry to calculate how much heat is released when a specific amount of it is burned. Let's start by working out the molar mass of acetone. The molar mass is the weight of one mole of a substance (in grams per mole), and it's as important to chemists as a teaspoon is to bakers. The molar mass of acetone \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}\) is \[58.08 \, \text{g/mol}\], which we get by adding the individual molar masses of carbon, hydrogen, and oxygen.

Back to the stoichiometry, we then use the formula \[\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\] to find out how many moles of acetone we have. This gives us the information we need to proceed to the next step - figuring out the heat released.
Enthalpy Changes
Now, chemistry often brings up the topic of energy, and this is where enthalpy changes come into play. Every chemical reaction involves energy. The term 'enthalpy' is a measure of the total energy of a thermodynamic system, and an 'enthalpy change' is the difference in energy between the reactants and the products in a reaction.

In the context of our reaction - the combustion of acetone - enthalpy change is represented as \(\Delta H_{\text{nxn}}\). The negative sign indicates that energy is being released, as heat, into the surroundings. This makes it an exothermic reaction. Exothermic reactions are like energetic movie blockbusters - they release energy, creating heat and sometimes light.

The enthalpy change for the combustion of one mole of acetone is \[-1790 \, \text{kJ}\]. Using our previously calculated number of moles of acetone, we can find the total heat released by multiplying by this enthalpy change, giving us a grand total using the equation \[\text{total heat released} = \text{number of moles} \times \Delta H_{\text{nxn}}\].

By understanding the amount of energy released, we can make connections to real-world applications, such as the energy content of fuels. Remember, enthalpy is more than just a number; it's a storyteller, narrating how substances interact energetically.

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Most popular questions from this chapter

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. $$ 2 \mathrm{~K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{KCl}(s) $$ (a) \(1 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(1.8 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (c) \(2.2 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (d) \(14.6 \mathrm{~mol} \mathrm{~K} ; 7.8 \mathrm{~mol} \mathrm{Cl}_{2}\)

For each reaction, calculate how many moles of the product form when \(0.112 \mathrm{~mol}\) of the reactant in color completely reacts. Assume there is more than enough of the other reactant. (a) \(2 \mathrm{Ca}(\mathrm{s})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CaO}(s)\) (b) \(4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(4 \mathrm{~K}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~K}_{2} \mathrm{O}(s)\) (d) \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\)

Determine the theoretical yield of \(C\) when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react in the generic reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ (a) \(1 \mathrm{~mol} \mathrm{~A} ; 1 \mathrm{~mol} \mathrm{~B}\) (b) \(2 \mathrm{~mol} \mathrm{~A} ; 2 \mathrm{~mol} \mathrm{~B}\) (c) \(1 \mathrm{~mol} \mathrm{~A} ; 3 \mathrm{~mol} \mathrm{~B}\) (d) \(32 \mathrm{~mol} \mathrm{~A} ; 68 \mathrm{~mol} \mathrm{~B}\)

Consider the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{rxn}}=-125 \mathrm{~kJ} $$ Determine the amount of heat emitted when each amount of reactant completely reacts (assume that there is more than enough of the other reactant). (a) \(2 \mathrm{~mol} \mathrm{~A}\) (b) \(3 \mathrm{~mol} \mathrm{~A}\) (c) \(3 \mathrm{~mol} \mathrm{~B}\) (d) \(5 \mathrm{~mol} \mathrm{~B}\)

Consider the generic reaction between reactants \(\mathrm{A}\) and \(\mathrm{B}\) : $$ 3 \mathrm{~A}+4 \mathrm{~B} \longrightarrow 2 \mathrm{C} $$ If a reaction vessel initially contains \(9 \mathrm{~mol} \mathrm{~A}\) and \(8 \mathrm{~mol} \mathrm{~B}\), how many moles of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) will be in the reaction vessel after the reactants have reacted as much as possible? (Assume \(100 \%\) actual yield.)

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